If `(a_(2)a_(3))/(a_(1)a_(4))=(a_(2)+a_(3))/(a_(1)+a_(4))=3((a_(2)-a_(3))/(a_(1)-a_(4)))`, then `a_(1),a_(2),a_(3),a_(4)` are in

Options

A. AP
B. GP
C. HP
D. none of these

Explanation

To solve the problem, we need to analyze the given equation:

\[ \frac{a_2 a_3}{a_1 a_4} = \frac{a_2 + a_3}{a_1 + a_4} = 3 \cdot \frac{a_2 - a_3}{a_1 - a_4} \]

Let's denote:

1. \( x = \frac{a_2 a_3}{a_1 a_4} \)
2. \( y = \frac{a_2 + a_3}{a_1 + a_4} \)
3. \( z = 3 \cdot \frac{a_2 - a_3}{a_1 - a_4} \)

From the equation, we have:

\[ x = y = z \]

### Step 1: Equating \( x \) and \( y \)

Starting with \( x = y \):

\[ \frac{a_2 a_3}{a_1 a_4} = \frac{a_2 + a_3}{a_1 + a_4} \]

Cross-multiplying gives:

\[ a_2 a_3 (a_1 + a_4) = a_1 a_4 (a_2 + a_3) \]

Expanding both sides:

\[ a_2 a_3 a_1 + a_2 a_3 a_4 = a_1 a_4 a_2 + a_1 a_4 a_3 \]

Rearranging terms:

\[ a_2 a_3 a_1 - a_1 a_4 a_2 + a_2 a_3 a_4 - a_1 a_4 a_3 = 0 \]

Factoring out common terms:

\[ a_2 (a_3 a_1 - a_1 a_4) + a_4 (a_2 a_3 - a_1 a_3) = 0 \]

### Step 2: Equating \( y \) and \( z \)

Now, equating \( y \) and \( z \):

\[ \frac{a_2 + a_3}{a_1 + a_4} = 3 \cdot \frac{a_2 - a_3}{a_1 - a_4} \]

Cross-multiplying gives:

\[ (a_2 + a_3)(a_1 - a_4) = 3(a_2 - a_3)(a_1 + a_4) \]

Expanding both sides:

\[ a_2 a_1 - a_2 a_4 + a_3 a_1 - a_3 a_4 = 3(a_2 a_1 + a_2 a_4 - a_3 a_1 - a_3 a_4) \]

Rearranging terms:

\[ a_2 a_1 - 3 a_2 a_1 - a_2 a_4 + 3 a_2 a_4 + a_3 a_1 - 3 a_3 a_1 - a_3 a_4 + 3 a_3 a_4 = 0 \]

Combining like terms:

\[ -2 a_2 a_1 + 2 a_2 a_4 - 2 a_3 a_1 + 2 a_3 a_4 = 0 \]

Factoring out common terms:

\[ 2(-a_2 a_1 + a_2 a_4 - a_3 a_1 + a_3 a_4) = 0 \]

### Step 3: Conclusion

From the equations derived, we can conclude that:

\[ \frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \frac{1}{a_4} \]

are in Arithmetic Progression (AP). Therefore, \( a_1, a_2, a_3, a_4 \) are in Harmonic Progression (HP).

### Final Answer

Thus, \( a_1, a_2, a_3, a_4 \) are in **Harmonic Progression (HP)**.

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